[PHP-DEV] PHP 4.0 Bug #8860 Updated: variable array not working
ID: 8860
Updated by: cynic
Reported By: [EMAIL PROTECTED]
Old-Status: Open
Status: Closed
Bug Type: Scripting Engine problem
Assigned To:
Comments:
no. it will echo contents of $three. if you don't have that variable in current scope,
it'll echo null, and, depending on your settings, emit a warning.
Previous Comments:
---
[2001-01-23 09:57:18] [EMAIL PROTECTED]
according to Zeev this should work:
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
But with it doesn't
Greetz,
Wico
---
Full Bug description available at: http://bugs.php.net/?id=8860
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[PHP-DEV] PHP 4.0 Bug #8860 Updated: variable array not working
ID: 8860
User Update by: [EMAIL PROTECTED]
Status: Closed
Bug Type: Scripting Engine problem
Description: variable array not working
ok and that isn;t working either...
althoug i liked the first more (and still think it should be that way)
so this doesn't work either:
$three = "G";
echo "Test";
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
Previous Comments:
---
[2001-01-23 09:59:43] [EMAIL PROTECTED]
no. it will echo contents of $three. if you don't have that variable in current scope,
it'll echo null, and, depending on your settings, emit a warning.
---
[2001-01-23 09:57:18] [EMAIL PROTECTED]
according to Zeev this should work:
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
But with it doesn't
Greetz,
Wico
---
Full Bug description available at: http://bugs.php.net/?id=8860
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PHP Development Mailing List http://www.php.net/
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
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[PHP-DEV] PHP 4.0 Bug #8860 Updated: variable array not working
ID: 8860
Updated by: cynic
Reported By: [EMAIL PROTECTED]
Status: Open
Bug Type: Scripting Engine problem
Assigned To:
Comments:
ah, sorry, should've read the report more carefully.
Previous Comments:
---
[2001-01-23 10:07:21] [EMAIL PROTECTED]
changed status
---
[2001-01-23 10:06:24] [EMAIL PROTECTED]
ok and that isn;t working either...
althoug i liked the first more (and still think it should be that way)
so this doesn't work either:
$three = "G";
echo "Test";
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
---
[2001-01-23 09:59:43] [EMAIL PROTECTED]
no. it will echo contents of $three. if you don't have that variable in current scope,
it'll echo null, and, depending on your settings, emit a warning.
---
[2001-01-23 09:57:18] [EMAIL PROTECTED]
according to Zeev this should work:
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
But with it doesn't
Greetz,
Wico
---
Full Bug description available at: http://bugs.php.net/?id=8860
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Re: [PHP-DEV] PHP 4.0 Bug #8860 Updated: variable array not working
At 15:20 23-1-01 +, [EMAIL PROTECTED] wrote:
ID: 8860
Updated by: cynic
Reported By: [EMAIL PROTECTED]
Status: Open
Bug Type: Scripting Engine problem
Assigned To:
Comments:
ah, sorry, should've read the report more carefully.
So what should it echo?
three or Grrr?
i think three
Greetz,
Wico
Previous Comments:
---
[2001-01-23 10:07:21] [EMAIL PROTECTED]
changed status
---
[2001-01-23 10:06:24] [EMAIL PROTECTED]
ok and that isn;t working either...
althoug i liked the first more (and still think it should be that way)
so this doesn't work either:
$three = "G";
echo "Test";
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var;
/* should echo "three" */
---
[2001-01-23 09:59:43] [EMAIL PROTECTED]
no. it will echo contents of $three. if you don't have that variable in
current scope, it'll echo null, and, depending on your settings, emit a
warning.
---
[2001-01-23 09:57:18] [EMAIL PROTECTED]
according to Zeev this should work:
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var;
/* should echo "three" */
But with it doesn't
Greetz,
Wico
---
Full Bug description available at: http://bugs.php.net/?id=8860
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PHP Development Mailing List http://www.php.net/
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
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[PHP-DEV] PHP 4.0 Bug #8860 Updated: variable array not working
ID: 8860
Updated by: stas
Reported By: [EMAIL PROTECTED]
Status: Open
Old-Bug Type: Scripting Engine problem
Bug Type: Feature/Change Request
Assigned To:
Comments:
It is definitely not meant to work this way. 'test[2]' is
not a variable name. $test is the variable (array), and
$test[2] is second element of the array. So it belongs to
feature requests.
Previous Comments:
---
[2001-01-23 10:20:59] [EMAIL PROTECTED]
ah, sorry, should've read the report more carefully.
---
[2001-01-23 10:07:21] [EMAIL PROTECTED]
changed status
---
[2001-01-23 10:06:24] [EMAIL PROTECTED]
ok and that isn;t working either...
althoug i liked the first more (and still think it should be that way)
so this doesn't work either:
$three = "G";
echo "Test";
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
---
[2001-01-23 09:59:43] [EMAIL PROTECTED]
no. it will echo contents of $three. if you don't have that variable in current scope,
it'll echo null, and, depending on your settings, emit a warning.
---
[2001-01-23 09:57:18] [EMAIL PROTECTED]
according to Zeev this should work:
$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should
echo "three" */
But with it doesn't
Greetz,
Wico
---
The remainder of the comments for this report are too long. To view the rest of the
comments, please view the bug report online.
Full Bug description available at: http://bugs.php.net/?id=8860
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